Problem: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $22.2$ years; the standard deviation is $3.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living less than $12.3$ years.
Answer: $22.2$ $18.9$ $25.5$ $15.6$ $28.8$ $12.3$ $32.1$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $22.2$ years. We know the standard deviation is $3.3$ years, so one standard deviation below the mean is $18.9$ years and one standard deviation above the mean is $25.5$ years. Two standard deviations below the mean is $15.6$ years and two standard deviations above the mean is $28.8$ years. Three standard deviations below the mean is $12.3$ years and three standard deviations above the mean is $32.1$ years. We are interested in the probability of a porcupine living less than $12.3$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the porcupines will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $12.3$ years and the other half $({0.15\%})$ will live longer than $32.1$ years. The probability of a particular porcupine living less than $12.3$ years is ${0.15\%}$.